Re: [DexterBase] SL jumping
Bps:
> …It seems to me that this system would require a greater force to actually break the 80lb break cord…
> …As to how much force this is, I don't know…
Simply, twice as much force than the “traditional” set up of break cord.
Rfarris:
> I guess my uneducated reasoning saw the thicker rope taking most of the load, and eventually transferring it to the smaller cord.
No. it is NOT a matter of how thick is any rope/cord. It is a matter of how many “branches” there are to bear the load. Quite simple. 1 branch (=1 single line of rope/cord/whatever): full load; 1 loop (=2 branches): ½ load; 2 loops: ¼ load on each branch and so on. Then, in case of 1 loop, each branch takes ½ load: the system breaks where the one of the 2 branches is weaker (=the one that has got lower break load) than the other.
Dexter:
> My logic is that since spectra slings don't stretch at all, the load will be transferred to the breakcord.
Yes, the spectra sling streches very little (no system/no rope has got infinite rigidity), but that is NOT the point. The point is, as I said before, that the full load (=the force transmitted by the bridle) is divided by as many times as the number of branches of the system.
> It seems to me that since both ends of the static line sling are attached to the same loadpoint, that both sides must share equal amounts of the force.
Yes, this is true.
> Since the breakcord will part at 80 pounds, as soon as the system is loaded to that point, the breakcord will give. It's a weak link designed into the system.
Not at all. Yes, the loop of break cord will part at 80 lb - 36 kg, but to get to 80 lb - 36 kg on the loop of break cord, you must pull the bridle with a 160 lb - 72 kg force.
Can you see the scenario? You have a pulling force on the bridle of 160 lb - 72 kg (you are about to break the system): this force is transmitted entirely on the rapid link. Then, the force of 160 lb - 72 kg is divided by two simply because there are two branches after the rapid link towards the fixed point onto the structure: two branches forming the sling loop. Each branch of the sling takes (=must take, forces can neither disappear nor increase, they can only be distributed/shared among various parts of a static system) half of the load: so each of the branches of sling takes 80 lb - 36 kg. Now, there is one branch of the sling that contains the loop of break cord: yes, it is going to part, but at this stage you are exerting, through your body, harness, lines, canopy, bridle a force of 160 lb - 72 kg, that is exactly the double of the force at which the system is supposed to be loaded and break.
Be careful: in a static system where there are forces that are exerted along a pillar/rope/line/whatever, any time along such a system is inserted a “loop”, each branch of the loop takes half the load (the total force transmitted can neither vanish nor increase).
Remember that with a suitable system of pulleys/ropes/turns of the rope along/around the pulleys, if you anchor such a system on a high point that takes all the load (=weight), exerting with your hands/arms a small force you can easily pull up a very heavy load: a grand piano, a safe, a car, whatever. The more load/weight you want to lift, the more pulleys and the more passes of rope through/around the pulleys need to be done.
Now, to take the good in your system, Dexter, you should simply use the break cord in the following way: you do one loop with a knot around the rapid link, then, off this loop/knot must come out only one single line of break cord (=the break cord pure and simple), then this single line of break cord “travels” all alone towards the loop in the sling: here you fix it with another loop/knot. This system is going to break when a force of 80 lb - 36 kg is exerted by the bridle onto the system.
Just after I wrote the above, I am wondering why you are use the rapid link, a component that is unnecessary (and that could fail…), at this stage. Has the sling got two (small) loops? You fix one end of sling with a larks head knot onto the bridle loop, then you fix the other end again onto the bridle loop using the break cord with a SINGLE LINE OF BREAK CORD (loop/knot, single line of break cord, loop/knot). Furthermore, the rapid link creates a sharp bend for the break cord: this way it could break at a load lower than the load it is supposed to break. We use static lines onto the structure to avoid sharp bends/sharp point of metallic structure to make the break cord properly, and then you introduce a very sharp bend through the rapid link (even if smooth surface, but still is very sharp bend (=small curvature radius))?!?!?!?
Just my 0.02€.